Where is the vertex of the parabola f(x) = x^2 - 5x + 6?

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In summary, the conversation discusses an equation, f(x) = x^2 - 5x + 6, and its vertex. It is mentioned that the points 2 and 3 on the x axis reach 0, and the question is asked about the location of the vertex. The speaker provides two methods for finding the vertex, one using the fact that the vertex is halfway between two points of equal height and the other using the vertex form of a parabola. The conversation also briefly mentions calculus and the speaker asks the other person about their knowledge on the subject.
  • #1
darkfire313
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See, there is an equation which is f(x) = x^2 - 5x + 6. On the points 2 and 3 of the x axis, they reach 0. So where is the vertex? I'm guessing its -.25, but just to make sure i had to ask. Well, our class seemed to have a lot of it and our teacher, Ms. Knudsvig, didn't know.
 
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  • #2
darkfire313 said:
See, there is an equation which is f(x) = x^2 - 5x + 6. On the points 2 and 3 of the x axis, they reach 0. So where is the vertex? I'm guessing its -.25, but just to make sure i had to ask. Well, our class seemed to have a lot of it and our teacher, Ms. Knudsvig, didn't know.
You mean that f(2)= 4- 10+ 6= 0 and f(3)= 9- 15+ 6= 0. Because this parabola has vertical axis, it vertex is exactly half way between 2 and 3, at 5/2. f(5/2)= 25/4- 25/2+ 6= 25/4- 50/4+ 24/4= -1/4= 0.25.

You could also get that, without using the fact that the vertex is exactly halfway between to points of equal height, by completing the square:

5/2= 5/2 and (5/2)^2= 25/4 so f(x)= x^2- 5x+ 25/4- 25/4+ 6= (x- 5/2)^2- 25/4+ 24/4= (x- 5/2)^2- 1/4.

Since a square is never negative, that is always -1/4 plus something. When x= 5/2, (5/2- 5/2)^2- 1/4= -1/4. The lowest point, the vertex, is (5/2, -1/4).
 
  • #3
darkfire313 said:
See, there is an equation which is f(x) = x^2 - 5x + 6. On the points 2 and 3 of the x axis, they reach 0. So where is the vertex?

To find the vertex, all you have to do is write the equation in vertex form. Standard from for a parabola is ax^2+bx+c. Vertex form for a parabola is y= a(x-h)^2+k. Once you have the equation in this form, h and k are the coordinates for your vertex.
 
  • #4
How much do you (darkfire) know about calculus?
 

1. What is a parabola?

A parabola is a type of curve that is shaped like a U. It is a conic section, meaning it is formed by the intersection of a plane and a cone. In mathematics, parabolas are described by quadratic equations and have many real-world applications, such as in physics and engineering.

2. How do you graph a parabola?

To graph a parabola, you will need to plot at least three points on the curve. These points can be found by substituting different values for x into the parabola's equation and solving for y. Once you have three points, you can connect them with a smooth curve to create the parabola.

3. What is the focus and directrix of a parabola?

The focus of a parabola is a fixed point on the axis of symmetry, and the directrix is a line perpendicular to the axis of symmetry. The distance between the focus and any point on the parabola is equal to the distance between that point and the directrix. These properties are essential in defining and understanding the shape of a parabola.

4. How do you find the vertex of a parabola?

The vertex of a parabola is the point where the parabola changes direction, or where the slope of the curve is zero. To find the vertex, you can use the formula x = -b/2a, where a and b are the coefficients of the quadratic equation in standard form, y = ax^2 + bx + c. The y-coordinate of the vertex can be found by substituting the x-coordinate into the equation.

5. How are parabolas used in real life?

Parabolas have many real-world applications, such as in physics to describe the trajectory of projectiles and in engineering to design bridges and buildings. They are also used in optics for reflecting light and in satellite dishes to focus signals. Additionally, parabolas can be used to model the motion of objects under the influence of gravity, making them essential in understanding and studying physics and motion.

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